Q33E In Section 4.3, it is shown that... [FREE SOLUTION]

Chapter 4: Q33E (page 136)

In Section 4.3, it is shown that $唯 {(x,t)=Ae}^{颈(办虫-蝇 t)}$ satisfies the free-particle Schr枚dinger equation for all $x$ and $t$ , provided only that the constants are related by ${(鈩� k)}^{2} / 2m = 鈩� 蝇$ . Show that when the function, $唯 (x,t)=Acos(kx 鈭� 0x)$ is plugged into the Schrodinger equation, the result cannot possibly hold for all values of x and t, no matter how the constants may be related.

Short Answer

Expert verified

We can't come up with a condition on $蝇$ or k that will make the proposed solution valid in this situation under any circumstances.

Step by step solution

Step 1:Schrodinger equation.

So, let's take the above function and plug it back into the Schrodinger equation to see if we can come up with a condition on the value of $蝇$ or $k$ that will allow us to get this result.

$鈭� \frac{{鈩�}^{2}}{2 m} \frac{{鈭�}^{2} 唯}{鈭� x^{2}} = i 鈩� \frac{鈭� 唯}{鈭� t}$ 鈥︹赌︹赌︹赌︹赌︹赌�(1)

Schrodinger equation for the given wavefunction

The wavefunction given is

$唯 = Acos(kx-蝇t)$ 鈥︹赌︹赌︹赌︹赌︹赌�.(2)

Substitute the value of wavefunction from equation (2) into equation (1), and we get,

$\begin{array}{l} \frac{{鈭�}^{2} 唯}{鈭� x^{2}} = 鈭� {Ak}^{2} cos(kx-蝇t) \\ \frac{鈭� 唯}{鈭� t} = A蝇sin(kx-蝇t) \end{array}$

$\begin{array}{l} 鈭� \frac{{鈩�}^{2}}{2m} 脳 ({-Ak}^{2} cos(kx-蝇t)) = i 鈩� 脳 A蝇sin(kx-蝇t) \\ \frac{鈩� k^{2}}{2m} = i蝇tan(kx-蝇t) \end{array}$

Conclusion.

Unlike the exponential case, this final expression has a function that depends on both position and time, but the left-hand side is constant. As a result, we can't discover a condition on $k$ or $蝇$ that will make the two sides equal, and the sine function as a solution is no different.